A uniform elastic plank moves due to a constant force `F_(0)` applied at one end whose area is `S`. The Young's modulus of the plank is `Y`. The strain produced in the direction of force is

Options

A. `F_(o)/(2SY)`
B. `F_(o)/(SY)`
C. `(2F_(0))/(SY)`
D. `sqrt(2F_(0))/(SY)`

Explanation

To find the strain produced in the direction of the force on a uniform elastic plank subjected to a constant force \( F_0 \), we can follow these steps:

### Step-by-Step Solution:

1. **Identify the Given Data:**
- Applied force, \( F_0 \)
- Cross-sectional area, \( S \)
- Young's modulus, \( Y \)
- Length of the plank, \( L \)

2. **Define the Mass per Unit Length:**
- Let the total mass of the plank be \( M \).
- The mass per unit length of the plank is given by:
\[ \text{Mass per unit length} = \frac{M}{L} \]

3. **Consider an Element of the Plank:**
- Take a small element of length \( dx \) at a distance \( x \) from the starting point \( A \).
- The mass of the small element \( AC \) can be expressed as:
\[ \text{Mass of } AC = \left(\frac{M}{L}\right) \cdot dx \]

4. **Apply Newton's Second Law:**
- The net force acting on the small element \( AC \) is given by:
\[ F_0 - T = \text{mass of } AC \cdot \text{acceleration} \]
- The acceleration \( a \) of the plank can be expressed as:
\[ a = \frac{F_0}{M} \]
- Thus, we can write:
\[ F_0 - T = \left(\frac{M}{L}\right) \cdot dx \cdot \frac{F_0}{M} \]

5. **Express Tension \( T \):**
- Rearranging the equation gives:
\[ T = F_0 - \left(\frac{F_0}{L}\right) \cdot x \]

6. **Calculate the Change in Length \( \Delta L \):**
- The small change in length \( d\Delta L \) for the small element \( dx \) is given by:
\[ d\Delta L = \frac{T \cdot dx}{S \cdot Y} \]
- Substituting for \( T \):
\[ d\Delta L = \frac{\left(F_0 - \frac{F_0}{L} \cdot x\right) \cdot dx}{S \cdot Y} \]

7. **Integrate to Find Total Change in Length \( \Delta L \):**
- Integrate \( d\Delta L \) from \( 0 \) to \( L \):
\[ \Delta L = \int_0^L \frac{F_0 \cdot dx}{S \cdot Y} - \int_0^L \frac{F_0 \cdot x \cdot dx}{S \cdot Y \cdot L} \]
- The first integral gives:
\[ \frac{F_0 \cdot L}{S \cdot Y} \]
- The second integral evaluates to:
\[ \frac{F_0}{2 \cdot S \cdot Y} \]
- Therefore, the total change in length is:
\[ \Delta L = \frac{F_0 \cdot L}{2 \cdot S \cdot Y} \]

8. **Calculate Strain:**
- Strain \( \epsilon \) is defined as the change in length per unit original length:
\[ \epsilon = \frac{\Delta L}{L} = \frac{F_0}{2 \cdot S \cdot Y} \]

### Final Result:
The strain produced in the direction of the force is:
\[ \epsilon = \frac{F_0}{2 \cdot S \cdot Y} \]